# 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2020 Canadian Senior Mathematics Contest Solutions

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2020 Canadian Senior Mathematics Contest Solutions

(a) When $p=33$ and $q=216$,
$$f(x)=x^3-33 x^2+216 x=x\left(x^2-33 x+216\right)=x(x-9)(x-24)$$
since $9+24=33$ and $9 \cdot 24=216$, and
$$g(x)=3 x^2-66 x+216=3\left(x^2-22 x+72\right)=3(x-4)(x-18)$$
since $4+18=22$ and $4 \cdot 18=72$.
Therefore, the equation $f(x)=0$ has three distinct integer roots (namely $x=0, x=9$ and $x=24$ ) and the equation $g(x)=0$ has two distinct integer roots (namely $x=4$ and $x=18$ ).
(b) Suppose first that the equation $f(x)=0$ has three distinct integer roots.
Since $f(x)=x^3-p x^2+q x=x\left(x^2-p x+q\right)$, then these roots are $x=0$ and the roots of the quadratic equation $x^2-p x+q=0$ which are
$$x=\frac{p \pm \sqrt{p^2-4(1) q}}{2(1)}=\frac{p \pm \sqrt{p^2-4 q}}{2}$$
For the roots of $x^2-p x+q=0$ to be distinct, we need $p^2-4 q$ to be positive.
For the roots of $x^2-p x+q=0$ to be integers, we need each of $p \pm \sqrt{p^2-4 q}$ to be an integer, which means that $\sqrt{p^2-4 q}$ is an integer, which means that $p^2-4 q$ must be a perfect square.
Therefore, $p^2-4 q$ is a positive perfect square.
Suppose also that the equation $g(x)=0$ has two distinct integer roots.
The roots of the equation $3 x^2-2 p x+q=0$ are
$$x=\frac{2 p \pm \sqrt{(2 p)^2-4(3)(q)}}{2(3)}=\frac{2 p \pm \sqrt{4 p^2-12 q}}{6}=\frac{p \pm \sqrt{p^2-3 q}}{3}$$
As above, for these roots to be distinct, we need $p^2-3 q$ to be positive and a perfect square. Furthermore, since the roots of the equation $3 x^2-2 p x+q=0$ are distinct integers, then the roots of the equation $x^2-\frac{2 p}{3} x+\frac{q}{3}=0$ are also distinct integers.

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2020 Canadian Senior Mathematics Contest Solutions

(c) The goal of this solution is to show that there are infinitely many pairs of positive integers $(p, q)$ with certain properties. To do this, we do not have to find all pairs $(p, q)$ with these properties, as long as we still find infinitely many such pairs. This means that we can make some assumptions as we go. Rather than making all of these assumptions at the very beginning, we will add these as we go.

To begin, we assume that $p$ and $q$ are positive integers with $p$ a multiple of 3 and $q$ a multiple of 9 . (Assumption #1)
Thus, we write $p=3 a$ and $q=9 b$ for some positive integers $a$ and $b$.
Suppose that $a$ and $b$ have the additional property that $a^2-3 b=m^2$ and $a^2-4 b=n^2$ for some positive integers $m$ and $n$. (Assumption #2)
These first two Assumptions are not surprising given the results of (b).
In this case, the non-zero solutions of $f(x)=0$ are
$$x=\frac{p \pm \sqrt{p^2-4 q}}{2}=\frac{3 a \pm \sqrt{(3 a)^2-4(9 b)}}{2}=\frac{3 a \pm 3 \sqrt{a^2-4 b}}{2}=\frac{3 a \pm 3 n}{2}$$
and the solutions of $g(x)=0$ are
$$x=\frac{2 p \pm \sqrt{4 p^2-12 q}}{6}=\frac{p \pm \sqrt{p^2-3 q}}{3}=\frac{3 a \pm 3 \sqrt{a^2-3 b}}{3}=a \pm m$$
These solutions are all integers as long as the integers $3 a \pm 3 n$ are both even, which is equivalent to saying that $a$ and $n$ are both even or both odd (that is, have the same parity). Since $a^2-4 b=n^2$, this means that $a^2+n^2=4 b$, which is even, which means that $a^2$ and $n^2$ have the same parity, which means that $a$ and $n$ have the same parity.
Further, we note that since $p=3 a$ and $q=9 b$ then both $p$ and $q$ are divisible by 3 and so $\operatorname{gcd}(p, q)=3$ exactly when $a$ and $3 b$ have no further common divisors larger than 1 .
Therefore, to find an infinite number of pairs of positive integers $(p, q)$ which satisfy the given conditions, we can find an infinite number of pairs of positive integers $(a, b)$ for which $a^2-3 b$ and $a^2-4 b$ are both positive perfect squares, and where $\operatorname{gcd}(a, 3 b)=1$.

# 滑铁卢数学竞赛代考

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2020 Canadian Senior Mathematics Contest Solutions

(a) 当 $p=33$ 和 $q=216$ ，
$$f(x)=x^3-33 x^2+216 x=x\left(x^2-33 x+216\right)=x(x-9)(x-24)$$

$$g(x)=3 x^2-66 x+216=3\left(x^2-22 x+72\right)=3(x-4)(x-18)$$

(b) 首先假设方程 $f(x)=0$ 具有三个不同的整数根。

$$x=\frac{p \pm \sqrt{p^2-4(1) q}}{2(1)}=\frac{p \pm \sqrt{p^2-4 q}}{2}$$

$$x=\frac{2 p \pm \sqrt{(2 p)^2-4(3)(q)}}{2(3)}=\frac{2 p \pm \sqrt{4 p^2-12 q}}{6}=\frac{p \pm \sqrt{p^2-3 q}}{3}$$

## 数学竞赛代写|滑铁卢数学竞赛代写Waterloo Math Contest代考|2020 Canadian Senior Mathematics Contest Solutions

(c) 这个解的目的是证明有无穷多对正整数 $(p, q)$ 具有一定的属性。为此，我们不必找到所有对 $(p, q)$ 有了 这些性质，只要我们仍然找到无限多这样的对。这意味着我们可以随时做出一些假设。我们不会在一开 始就做出所有这些假设，而是在进行中添加这些假设。

$$x=\frac{p \pm \sqrt{p^2-4 q}}{2}=\frac{3 a \pm \sqrt{(3 a)^2-4(9 b)}}{2}=\frac{3 a \pm 3 \sqrt{a^2-4 b}}{2}=\frac{3 a \pm 3 n}{2}$$

$$x=\frac{2 p \pm \sqrt{4 p^2-12 q}}{6}=\frac{p \pm \sqrt{p^2-3 q}}{3}=\frac{3 a \pm 3 \sqrt{a^2-3 b}}{3}=a \pm m$$

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